# Modeling By Calculus?

1. Mar 8, 2006

### WARGREYMONKKTL

Modeling By Calculus?????!???

HI
I AM A STDENT WHO REALLY EXCITED OF PHYSICS. ESPECIALLY A BOUT THE MATHS THAT APPLIED IN PHYSICS.
SOMEBODY SAYS THAT WITH CALCULUS WE HAVE HAD A VERY POWERFUL TOOL FOR MODELING.
I HOPE MANY PEOPLE WITH DISCUSS ABOUT THAT IN THIS THREAD SO WE CAN KNOW MORE ABOUT MATH MODELING FOR PHYSICS.
EXAMPLE : HOW CAN WE APPLY DERIVATIVE IN PHYSICS TO FIND VELOCITY OF A MOTION, OR THE INTENSITY OF A CURRENT?

2. Mar 8, 2006

### neurocomp2003

umm pick up any ODE or PDE text or MATH Phys
A good text from teh math bio side of things is Lisa Keshet's...though i don't like the fact that the math is mixed in with the bio stuff...she does
organize the book well into ODE/PDE/DynSys/Bifurcation Theory

3. Mar 8, 2006

### AlphaNumeric

If something changes smoothly and it's rate of change is related to some other quantity of the system which also changes smoothly (such as electromagnetic fields, liquids, populations, gravity, temperature, almost everything in nature) then you can (in theory) model the system using differential equations.

Navier's equation for fluid flow :
$$(\lambda + 2 \mu) \nabla (\nabla . \mathbf{u}) - \mu \nabla \times (\nabla \times \mathbf{u}) = \rho \frac{\partial^{2} \mathbf{u}}{\partial t^{2}}$$

Schroedinger's equation :
$$i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}} + V(x)\psi$$

Those two are both partial differential equations which crop up in large amounts of applied mathematics.

4. Mar 12, 2006

### WARGREYMONKKTL

if you say so how can i model a phenomanon such as a turnado?
will i choose a particle and suggest that it is in side the turnado and use calculus model to set up an equation for that turnado????????
i don't get it.
thanks you

5. Mar 12, 2006

### WARGREYMONKKTL

what does those equations mean i don't get it.
i am at a very low level of math.

6. Mar 12, 2006

### HallsofIvy

Staff Emeritus
Then wait. No explanation will make sense until you have the math to understand the explanation.

As simply as possible, calculus allows us to define rate of change at a specific instant rather than just average rate of change. That's crucial to any science.

7. Mar 12, 2006

### arildno

The basic equation you'll met first in physics that shows the usefulness of calculus, is Newton's 2.law of motion, F=ma.
"a", the acceleration of the particle, is the second derivative of the position of the particle with respect to time.
Thus, if the forces F are known, you can find the position of the particle as a function of time.

8. Mar 12, 2006

### AlphaNumeric

Then you're bitting off more than you can chew in most cases. As HallsofIvy says, I'd explain them then you'd ask what my explainations meant because the notions would not be known to you.
There are models for tornados, but are extremely complex applications of fluid and thermodynamic equations. A tornado has many different things which affect it (heat, air pressure, humidity, proximity to land, global air currents) and the models we have are not enormously accurate, but getting better all the time.

By the way, fluids aren't (usually) modelled as a collection of particles, but a continous medium. Only in special cases do fluids get considered to have particle makeup.

A simple example of using a differential equation to model something in real life is nuclear decay of a sample of material like uranium. In this case the amount of radiation given off is proportional to the number of radioactive atoms, so you have

$$\frac{dN}{dt} \propto N$$

which gives you the differential equation (when putting in a constant of proportionality)

$$\frac{dN}{dt} = -kN$$

The minus is because the sample is shrinking. Solving gives

$$N = N_{0}e^{-kt}$$

Therefore radioactive material decays in an exponential fashion. The value of 'k' is different for each material, and relates to the 'half life', which is the commonly quoted quantity when talking about things which decay.

9. Mar 12, 2006

### WARGREYMONKKTL

no i don't understand if the sign look like the number 6 in front of t variables in the fluid equation that was provided. what does it mean?
eventhogh my math is not good but i can understand the meaning of the calculus you show.
i just don't understand the physiologically sign( "6" in front of the variable). what is the operation for that sign? i mean what is it mathematical meaning?

10. Mar 12, 2006

### WARGREYMONKKTL

thank you every much for your careful explanation!
as you talking a bout the radiation of uranium. i want to ask you a question.
we know that Uranium 235 is a radiactive substance. will its radiations make the environment( around the decaying unranium) become radiactive? is there an electromagnetic field around it? if there is a EM-F
what will happen when a photon get into it? you don't need to answer all this . i just reallly interesting in the nuclear physics.
thanks a gain!
<vincent>

11. Mar 12, 2006

### neurocomp2003

ah um its not a six hehe...its the PDE derivative sign( change of what ever variable follows teh sign)...if you don't know that sign then your very far from learning applied math.

as for your tornado equatoin as mentioned above you must decide what your smallest fundamental element in is...like in population dynamics the smalled entity is the individual..."N" would represent teh populatoin size and thus dN change in population.

THen you must also decide what physics model(type of motion/force) you will be using in sim. and then you build your eq'n accordingly

12. Mar 12, 2006

### WARGREYMONKKTL

CAN YOU TELL ME MORE ABOUT THAT?
I DONT'T GET IT.
THANK YOU!

13. Mar 12, 2006

### WARGREYMONKKTL

CAN YOU GIVE ME SOME INFORMATION ABOUT THAT PDE DERIVATIVE SIGN?
<I AM AN HIGH SCHOOL STUDENT>

14. Mar 12, 2006

### neurocomp2003

if you are a high school student are you taking calculus

www.mathworld.com everymath students best friend.

15. Mar 13, 2006

### WARGREYMONKKTL

thank you for the link you provide me.
can you please explain the meaning of the schrodinger's equation. how can we figure out the position of the object at $$psi(t,x)$$ i hope it working? a "little bit" confusing about it.